Problem: Simplify and expand the following expression: $ \dfrac{5t}{3t - 8}-\dfrac{2t - 8}{t + 6} $
In order to subtract expressions, they must have a common denominator. Get both fractions over a common denominator of $(3t - 8)(t + 6)$ Multiply the first term by $\dfrac{t + 6}{t + 6}$ $ \begin{align*} \dfrac{5t}{3t - 8} \times \dfrac{t + 6}{t + 6} & = \dfrac{(5t)(t + 6)}{(3t - 8)(t + 6)} \\ & = \dfrac{5t^2 + 30t}{(3t - 8)(t + 6)}\end{align*} $ Multiply the second term by $\dfrac{3t - 8}{3t - 8}$ $ \begin{align*} \dfrac{2t - 8}{t + 6} \times \dfrac{3t - 8}{3t - 8} & = \dfrac{(2t - 8)(3t - 8)}{(t + 6)(3t - 8)} \\ & = \dfrac{6t^2 - 40t + 64}{(t + 6)(3t - 8)}\end{align*} $ Now we have: $ = \dfrac{5t^2 + 30t}{(3t - 8)(t + 6)} - \dfrac{6t^2 - 40t + 64}{(t + 6)(3t - 8)} $ Now both terms have a common denominator we can subtract the numerators: $ = \dfrac{5t^2 + 30t - (6t^2 - 40t + 64)}{(3t - 8)(t + 6)} $ $ = \dfrac{5t^2 + 30t - 6t^2 + 40t - 64}{(3t - 8)(t + 6)} $ $ = \dfrac{-t^2 + 70t - 64}{(3t - 8)(t + 6)}$ Expand the denominator: $ = \dfrac{-t^2 + 70t - 64}{3t^2 + 10t - 48}$